\(n_{CaO}=\dfrac{6,72}{56}=0,12\left(mol\right)\)
CaO + H2O \(\rightarrow\) Ca(OH)2
de: 0,12 \(\rightarrow\) 0,12
b, \(m_{Ca\left(OH\right)_2}=0,12.74=8,88g\)
\(n_{CO_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
Ta co: \(\dfrac{0,12}{1}>\dfrac{0,08}{1}\) => Ca(OH)2 dư
Ca(OH)2 + CO2 \(\rightarrow\) CaCO3 + H2O
de: 0,08 \(\rightarrow\) 0,08
c, \(m_{CaCO_3}=0,08.100=8g\)